Let $k$ be an algebraically closed field, $A$ a finite dimensional (unital associative) $k$-algebra of finite dimension, and $G$ a torus over $k$ acting on $A$.

What does $k$-algebra $A \rtimes G$ stand for? More specifically how to define addition, multiplication, and scaler multiplication?

The only usage of $\rtimes$ that I know is that of semidirect product and trivial extension algebra. But none of them make sense here. I came across this in the proposition 5.1 in this paper (alternative link); the author says the statement is classic and doesn't explain what he meant by the symbol. Could you help me?

  • $\begingroup$ My guess (I don't really feel like reading through the arguments of a paper in French to see if I am right) is that it is the skew group algebra (see for example the book by Auslander-Reiren-Smalø for a reference). This is at least the most common thing to construct out of an algebra together with a group acting on it. $\endgroup$ – Tobias Kildetoft Dec 7 '15 at 7:32
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    $\begingroup$ I only know this notation for the crossed product en.wikipedia.org/wiki/Crossed_product $\endgroup$ – user197416 Dec 7 '15 at 7:40
  • $\begingroup$ I don't have access to this paper. Can you quote the relevant proposition? $\endgroup$ – Qiaochu Yuan Dec 7 '15 at 8:28
  • $\begingroup$ @QiaochuYuan I added an alternative arXiv link. $\endgroup$ – Orat Dec 7 '15 at 13:48

In a semidirect product $A\rtimes B$, the group $B$ acts on $A$. That is, there is a homomorphism from $B$ into $\operatorname{Aut}(A)$.

Here, you have $G$ acting on $A$, but in a different context. So maybe that is motivating using the notation $A\rtimes G$.

With groups, $A\rtimes B$ is the collection of pairs $(a,b)$, and there is a new group operation $(a_1,b_1)(a_2,b_2)=(a_1a_2^{b_1},b_1b_2)$ where $a_2^{b_1}$ is the image of $a_2$ under the action from $b_1$.

In OP's setting $A\rtimes G$ is the collection of pairs $(a,g)$, with $k(a,g)=(ka,g)=(a,kg)$. You again can have $(a_1,g_1)(a_2,g_2)=(a_1a_2^{g_1},g_1g_2)$, making $A\rtimes G$ an algebraic group. I'm not sure you can get an algebra out of this though, since the $G$ has no addition.

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    $\begingroup$ This does not really seem to answer the question. $\endgroup$ – Tobias Kildetoft Dec 7 '15 at 7:58
  • $\begingroup$ @TobiasKildetoft Oh I see. I thought the product structure was clear but that OP was asking why use this notation. I'll edit. $\endgroup$ – alex.jordan Dec 7 '15 at 7:59
  • $\begingroup$ I am not convinced the the edit has the correct structure. There are certainly ways to construct algebras from a group acting on an algebra (though the skew group algebra I mentioned is usually for finite groups acting on artin algebras, and I am not sure how far it generalizes). $\endgroup$ – Tobias Kildetoft Dec 7 '15 at 8:10
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    $\begingroup$ This is incorrect. If $G$ were just a group acting on an algebra $A$, then $A \rtimes G$ (which you might call, depending on taste, the semidirect product, the crossed product, or the skew group algebra...) would denote the algebra (not group) given by adjoining to $A$ an element $g$ for every $g \in G$ satisfying the relations in $G$ and the additional relations $grg^{-1} = \varphi(g) r$ where $\varphi(g)$ is the action. What's confusing about the OP's question is that $G$ is an algebraic group, not just a bare group. $\endgroup$ – Qiaochu Yuan Dec 7 '15 at 8:26
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    $\begingroup$ @alex: I disagree. In fact the problem is that the semidirect product is in no sense a product. It's an example of a homotopy quotient. $\endgroup$ – Qiaochu Yuan Dec 7 '15 at 17:46

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